In a water pistol, a piston drives water through a larger...

In a water pistol, a piston drives water through a larger tube of radius 1.00 $\mathrm{cm}$ into a smaller tube of radius 1.00 $\mathrm{mm}$ as in Figure $\mathrm{P9} .41 .$ (a) If the pistol is fired horizontally at a height of $1.50 \mathrm{m},$ use ballistics to determine the time it takes water to travel from the nozale to the ground. (Neglect air resistance and assume atmospheric pressure is 1.00 atm. (b) If the range of the stream is to be 8.00 $\mathrm{m}$ , with what speed must the stream leave the nozale? (c) Given the areas of the nozzle and cylinder, use the equation of continuity to calculate the speed at which the plunger must be moved. (d) What is the pressure at the nozzle? (e) Use Bernoulli's equation to find the pressure needed in the larger cylinder. Can gravity terms be neglected? (f) Calculate the force that must be exerted on the trigger to achieve the desired range. (The force that must be exerted is due to pressure over and above atmospheric pressure.)

Key Concepts

Force from Pressure Differences

The force required to move the piston or trigger in a device like a water pistol is directly linked to the pressure difference acting over a specific area. This concept, often expressed as F = ?P × A, connects the excess pressure above atmospheric levels with the area over which the pressure acts, resulting in a net force that propels the fluid.

Projectile Motion

Projectile motion concerns the motion of an object under the influence of gravity, typically following a curved trajectory. In problems like the water pistol, it is used to analyze the time it takes for water to fall from a given height and the horizontal distance it covers when launched, by applying kinematic equations that account for acceleration due to gravity.

Equation of Continuity

The equation of continuity is a principle in fluid dynamics which states that for an incompressible fluid, the mass flow rate must remain constant throughout its flow. This links the velocity of the fluid to cross-sectional areas, ensuring that when water flows from a larger area to a smaller area, its speed increases proportionally to conserve mass.

Bernoulli's Equation

Bernoulli's equation is an energy conservation principle in fluid dynamics that relates the pressure, kinetic energy per unit volume (due to the fluid's velocity), and potential energy per unit volume (due to elevation) along a streamline. It is instrumental in determining the pressure differences and flow speeds in systems where fluid is accelerated, such as a water pistol.

Pressure in Fluids

Pressure in fluid dynamics is defined as the force exerted per unit area and is a fundamental concept in understanding how fluids behave under various conditions. It plays a key role in determining the forces that drive fluid flow, and in systems like the water pistol, calculating the pressure at different sections is crucial to understanding the resulting speeds and forces.

Transcript

00:02 So this question combines the equations of fluid motion with the equations for projectile motion.

00:11 We're ultimately trying to calculate the force that must be applied to the trigger of a water pistol in order to achieve the desired results.

00:19 They actually give us the calculation we need to make so we can just follow their suggestions and it will lead us through the required calculations.

00:30 Now the first thing it asks us to find is how long it takes for water to fall from a height of 1 .5 meters if the water pistol is fired horizontally.

00:43 Well to do that we can do the equations for projectile motion.

00:46 We know that the vertical motion of a projectile, which the water is in this case, is going to be given by the vertical displacement delta y is equal to the initial velocity in the times the time plus 1 .5 times the acceleration in the y direction times the time squared.

01:08 That's for the vertical motion.

01:11 Now in this problem they said our water pistol is being fired horizontally so the water initially has no vertical velocity so that's zero and they said that we could ignore things like air resistance and so the acceleration of the water is just the acceleration of gravity which is a negative g so we can say that delta y is going to equal 1 .5 times a negative g times t squared.

01:45 We can now solve this equation for t as follows.

01:49 We multiply both sides by 2 that gives us 2 delta y and we'll divide both sides by a negative g that gives us 2 delta y over a negative g is going to be equal to t squared.

02:04 Taking the square root of both sides of the equation gives us the square root of 2 times delta y divided by a negative g equals t.

02:21 We can now plug in the numbers which we know.

02:25 Delta y is a negative 1 .5 meters.

02:29 That's the height from which the water pistol was fired so this becomes the square root of 2 times a negative 1 .5 and i'm going to leave off the units for clarity divided by a negative 9 .8.

02:49 We'll use that for g today.

02:51 The negatives then divide out leaving us with 2 times 1 .5 which is 3 divided by 9 .8 and square root of that so the square root of 2 times 1 square root of 2 times 1 .5 divided by 9 .8 gives us the value of 0 .55328.

03:18 So that is our time 0 .55328.

03:29 Now looking at the numbers they gave us in this problem we can reliably talk about three significant digits so we could say that this is equal to 0 .553 seconds and that's our answer for the first question.

03:49 However i'm going to use more significant digits than just that.

03:52 I'm going to use the longer answer there for later calculations so i don't over round.

03:58 The next question is if the range of the stream is to be 8 meters what is the speed must the stream have when it leaves the nozzle? well this is for part a.

04:12 We know that in the horizontal direction of a projectile it's moving at a constant velocity and so the distance that it travels in the horizontal direction should be that horizontal velocity times the time.

04:29 The time we've just calculated therefore we can just say that the distance it has to travel horizontally divided by that time is going to give us the velocity which we need.

04:40 And so then we're going to take the distance of 8 meters and we'll divide that by 0 .55328 seconds and 8 divided by 0 .55328 is equal to 14 .4592.

05:12 So therefore the speed that the water needs to be going is this 14 .4592.

05:23 Again i can round that off to three significant digits for the answer so the answer is going to be 14 .5 meters per second.

05:39 However again i'm going to use the longer answer there in later calculations so i don't over round.

05:46 Now question c then asks given the area of the nozzle and cylinder use the equation of continuity to calculate the speed at which the plunger must be moved.

05:58 Well the speed of the plunger moving must be the speed that the fluid is moving on the plunger side of the system.

06:06 From the equation of continuity we know that the area of the cylinder times the speed of the fluid in the cylinder is equal to the area of the nozzle times the speed of the fluid in the nozzle which we have just figured out.

06:22 That's the 14 .4592 meters per second that we just now figured out.

06:29 So therefore we're trying to find the velocity of the fluid in the cylinder.

06:34 The velocity in the cylinder must be the area of the nozzle divided by the area of the cylinder times the speed of the fluid in the nozzle.

06:47 Now these are given to be cylindrical shapes here so we know that the area of the nozzle is going to be pi times the radius of the nozzle squared.

06:59 Similarly the area of the cylinder is going to be pi times the radius of the cylinder squared and that then times the velocity of the nozzle.

07:10 So the pi's can divide out and this is going to be equal to then the radius of the nozzle divided by the radius of the cylinder squared times the velocity of the nozzle will give us the velocity of the cylinder.

07:32 Now we are told that the radius of the nozzle is one millimeter and the radius of the cylinder is one centimeter so they have a ratio of one to ten.

07:44 So this becomes one over ten squared times the velocity in the nozzle so that becomes one hundredth of the velocity of the nozzle.

07:58 So the velocity in the cylinder is equal to this is now 14 .4592 meters per second divided by a hundred.

08:15 That's easy enough to do that math in our head.

08:18 We can simply say the velocity of the cylinder, just move the decimal point two places, is 0 .144592 meters per second.

08:33 Again to give the answer that they might want for the text we need to round that off to three significant figures so this would be 0 .145 meters per second.

08:49 That would be the answer for this part...

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